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 Please help on Sql query
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Posted on 07-03-19 5:11 PM     Reply [Subscribe]
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​Hello , I have a table called Student consisting of Columns Name, and DateOfBirth. I would like to create a query Which Selects all Name whose DateOfBirth is on the same day. The DateOfBirth Column datatype is in DateTime format. I want the result in the Date format. In my table below I want row 1,2, and 7 for day 1 and 3,4 for day 2 as a result of the query.


 
Posted on 07-03-19 6:36 PM     [Snapshot: 49]     Reply [Subscribe]
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Use "GROUP BY" on DOB and get list of names in one column by using STRING AGG functions.
 
Posted on 07-03-19 6:38 PM     [Snapshot: 52]     Reply [Subscribe]
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select DATE_OF_BIRTH, LISTAGG(NAME, ',') WITHIN GROUP (ORDER BY NAME) AS NAME
from TABLE
GROUP BY DATE_OF_BIRTH;

Good Luck !!!
 
Posted on 07-04-19 10:17 AM     [Snapshot: 162]     Reply [Subscribe]
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Hello basnyatt,

When I run the query it throws the error saying:
"The function 'ListAgg' may not have a WITHIN GROUP clause."
 
Posted on 07-05-19 10:21 AM     [Snapshot: 259]     Reply [Subscribe]
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what RDBMS are you using
 
Posted on 07-05-19 12:18 PM     [Snapshot: 312]     Reply [Subscribe]
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select x.*, rownum as day_num from (
select to_date((to_char(dateOfBirth, 'YYYY-MM-DD')),'YYYY-MM-DD') date_of_Birth, listagg(Name, ',') within group(order by Name) Name_of_students
from student
group by to_date((to_char(dateOfBirth, 'YYYY-MM-DD')),'YYYY-MM-DD')
) x;


this works for Oracle. If you are using mysql try using string_agg instead of listagg.
 
Posted on 07-06-19 1:21 PM     [Snapshot: 410]     Reply [Subscribe]
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Hello basnyatt,

I am using Microsoft SQL Server 2012.
 
Posted on 07-07-19 11:37 AM     [Snapshot: 489]     Reply [Subscribe]
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Hello raajkm,

I need the skript for Microsoft SQL Server. Please help me.


Last edited: 08-Jul-19 08:07 AM

 
Posted on 07-08-19 12:14 PM     [Snapshot: 642]     Reply [Subscribe]
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WELL i do not use sql server 2012 but you might use the query as;

select to_date((to_char(S1.dateOfBirth, 'YYYY-MM-DD')),'YYYY-MM-DD') date_of_Birth,
stuff ((select distinct ','+ Name from student s2 where s2.name=s1.name FOR XML PATH(' ')),1,1,' ') as name_of_students

from student s1 group by to_date((to_char(S1.dateOfBirth, 'YYYY-MM-DD')),'YYYY-MM-DD')
Last edited: 08-Jul-19 12:16 PM
Last edited: 08-Jul-19 12:17 PM

 


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